Mixed Solution and Equilibrium Challenge

Introduction

Equilibrium questions in the laboratory rarely start with neat concentrations at equilibrium. Instead you are given a stock solution labelled by mass percent and density, or two solutions that are poured together. Before any equilibrium reasoning is possible, every species must be expressed as a concentration in the final mixture. This page links concentration conversion to the reaction quotient Q, the tool that tells you which way a mixture will move.

Core explanation

Stage 1: convert to molarity. A solution labelled with mass fraction w and density ρ is most easily handled by taking a basis of 1 dm³ (1000 cm³). The mass of solution is 1000ρ grams (ρ in g/cm³), the mass of solute is w × 1000ρ, and dividing by the molar mass gives moles per dm³. This is the density-based conversion from earlier in the unit, used here as the first link in a chain.

Stage 2: account for mixing. When volumes V₁ and V₂ are mixed, the total volume is usually taken as V₁ + V₂ for dilute aqueous solutions. Every species in each solution is diluted: c(new) = c(old) × V(old)/V(total). A frequent error is to dilute only one of the two solutions or to forget that spectator ions are diluted as well. Moles are conserved on mixing; concentrations are not.

Stage 3: react anything that goes essentially to completion. If the mixture contains a strong acid and a strong base, or another reaction with a very large K, deal with that stoichiometrically first, using moles. Only the leftover amounts are used in the equilibrium stage.

Stage 4: evaluate Q. Q has exactly the same form as K but uses the current concentrations. Then compare:

- Q < K: the reaction proceeds forward, forming more products. - Q > K: the reaction proceeds in reverse. - Q = K: the mixture is already at equilibrium.

For a sparingly soluble salt, Q is the ion product. For PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq), Q = [Pb²⁺][Cl⁻]². If Q exceeds Ksp, the solution is supersaturated and a precipitate is expected; if Q is below Ksp, all the salt stays dissolved. Note that solids and the solvent do not appear in Q.

Formulae

c = (1000 × ρ × w)/M in mol/dm³ (ρ in g/cm³). c(mixed) = c × V/V(total). Q = Πcᵢ^νᵢ (products over reactants, each raised to its coefficient).

Step-by-step reasoning

1. Convert every stock solution into mol/dm³. 2. Find the total volume after mixing and dilute every species. 3. Carry out any complete reaction using moles, then return to concentrations. 4. Write the Q expression from the balanced equation. 5. Substitute and compare Q with K or Ksp, and state the direction.

Visual explanation

Draw a horizontal number line of the value of Q with K marked on it. Mixtures to the left of K move right, towards products; mixtures to the right move left. Mixing two solutions places the starting point somewhere on the line, and dilution slides it, sometimes across K.

Real-world analogy

Q is like checking your bank balance against a target before deciding to save or spend. The number you compare must be today's balance after all the latest deposits and withdrawals, just as Q must use concentrations after mixing and dilution, not the values printed on the original bottles.

Real-world example

Water-treatment chemists predict scale formation by comparing the ion product of calcium and carbonate ions in a blended water supply with the Ksp of calcium carbonate. Mixing a hard groundwater with a softer surface water changes both concentrations, and the blend may or may not deposit scale in pipes.

Why?

Why must concentrations be recalculated after mixing? K is defined in terms of the concentrations actually present in the final solution. Adding volume lowers every concentration, and because Q multiplies powers of concentrations, a dilution can change Q by a large factor, especially when coefficients are greater than one.

Common misconception

"If both solutions contain the ions needed, a precipitate must form." Precipitation happens only when the ion product exceeds Ksp. Dilute solutions can be mixed with no visible change at all.

Worked example

Question: 20.0 cm³ of 0.0100 mol/dm³ Pb(NO₃)₂ is mixed with 30.0 cm³ of 0.0200 mol/dm³ NaCl. Ksp(PbCl₂) is about 1.7 × 10⁻⁵ at 25 °C. Does a precipitate form?

Reasoning: Total volume = 50.0 cm³. [Pb²⁺] = 0.0100 × 20.0/50.0 = 4.00 × 10⁻³ mol/dm³. [Cl⁻] = 0.0200 × 30.0/50.0 = 1.20 × 10⁻² mol/dm³. Q = (4.00 × 10⁻³)(1.20 × 10⁻²)² = 5.8 × 10⁻⁷. This is well below Ksp.

Answer: Q < Ksp, so no precipitate of PbCl₂ forms.

Quick check

1. A solution of ethanoic acid is 6.00% by mass with density 1.007 g/cm³. What is its concentration in mol/dm³ (M = 60.05 g/mol)? Answer: 1000 × 1.007 × 0.0600 = 60.4 g per dm³, which is 60.4/60.05 ≈ 1.01 mol/dm³.

Exam focus

Examiners reward a clear statement of the total volume, the diluted concentration of each ion, the correct Q expression with powers, and a verbal conclusion comparing Q with K. Remember that the stoichiometric coefficient becomes a power, so [Cl⁻] is squared for PbCl₂.

Advanced insight

At higher ionic strength, activities replace concentrations and are smaller than the concentrations, so a simple ion-product calculation can overestimate the tendency to precipitate. Close to the boundary Q ≈ Ksp, supersaturated solutions can also persist for some time because forming the first crystal nuclei is kinetically slow.

Summary

Mixed-solution equilibrium problems are chains: convert mass fraction and density into molarity, dilute every species to the total volume, remove any complete reactions stoichiometrically, and then evaluate Q. Comparing Q with K predicts the direction of change, and comparing the ion product with Ksp predicts whether a precipitate forms.

Practice questions

1. 10.0 cm³ of 0.100 mol/dm³ AgNO₃ is mixed with 90.0 cm³ of 1.0 × 10⁻⁴ mol/dm³ NaCl. Ksp(AgCl) ≈ 1.8 × 10⁻¹⁰. Does AgCl precipitate? Answer: [Ag⁺] = 0.0100 and [Cl⁻] = 9.0 × 10⁻⁵ mol/dm³, so Q = 9.0 × 10⁻⁷, which exceeds Ksp; a precipitate forms. 2. A reaction A + B ⇌ C has K = 50. After mixing, [A] = 0.10, [B] = 0.20 and [C] = 2.0 mol/dm³. Which way does it move? Answer: Q = 2.0/(0.10 × 0.20) = 100, which is greater than K, so the reaction moves in reverse. 3. Why does doubling the volume of a mixture by adding water change Q for A + B ⇌ C but not for A ⇌ B? Answer: For A ⇌ B, both concentrations halve and the ratio is unchanged; for A + B ⇌ C, Q doubles because the denominator has two concentration terms. 4. Why are spectator ions such as nitrate left out of Q for PbCl₂ precipitation? Answer: They do not appear in the dissolution equilibrium, so they do not enter the equilibrium expression, although they still affect the total volume and ionic strength.